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The manufacturer of a new compact car claims that the car will average at least

ID: 3315701 • Letter: T

Question

The manufacturer of a new compact car claims that the car will average at least 35 miles per gallon in general highway driving. For 100 test runs, the car averaged 35.5 miles per gallon with a standard deviation of 4 miles per gallon. Suppose that the manufacturer tests the following hypotheses: H subscript 0 : mu equals 35 H subscript 1 : mu greater than 35 Consider the 10% significance level. What is the power of the test if the actual average fuel efficiency is 36 miles per gallon? a. 95.254% b. 99.086% c. 80.511% d. 56.749% e. 88.877%

Explanation / Answer

Given that,

Standard deviation, =4

Sample Mean, X =35.5

Null, H0: =35

Alternate, H1: >35

Level of significance, = 0.1

From Standard normal table, Z /2 =1.2816

Since our test is right-tailed

Reject Ho, if Zo < -1.2816 OR if Zo > 1.2816

Reject Ho if (x-35)/4/(n) < -1.2816 OR if (x-35)/4/(n) > 1.2816

Reject Ho if x < 35-5.1264/(n) OR if x > 35-5.1264/(n)

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Suppose the size of the sample is n = 100 then the critical region

becomes,

Reject Ho if x < 35-5.1264/(100) OR if x > 35+5.1264/(100)

Reject Ho if x < 34.48736 OR if x > 35.51264

Implies, don't reject Ho if 34.48736 x 35.51264

Suppose the true mean is 36

Probability of Type II error,

P(Type II error) = P(Don't Reject Ho | H1 is true )

= P(34.48736 x 35.51264 | 1 = 36)

= P(34.48736-36/4/(100) x - / /n 35.51264-36/4/(100)

= P(-3.7816 Z -1.2184 )

= P( Z -1.2184) - P( Z -3.7816)

= 0.1115 - 0.0001 [ Using Z Table ]

= 0.1114

For n =100 the probability of Type II error is 0.1114

power of the test = 1- beta

= 1-0.1114

= 0.8886

= 88.86%

= 88.877%

option :E